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4x^2+24x=20
We move all terms to the left:
4x^2+24x-(20)=0
a = 4; b = 24; c = -20;
Δ = b2-4ac
Δ = 242-4·4·(-20)
Δ = 896
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{896}=\sqrt{64*14}=\sqrt{64}*\sqrt{14}=8\sqrt{14}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{14}}{2*4}=\frac{-24-8\sqrt{14}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{14}}{2*4}=\frac{-24+8\sqrt{14}}{8} $
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